V-Notch

Hydraulics · FE Reference Handbook section

Hydraulics

8 formulas

10 exam-style examples

~60 min

All Hydraulics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers V-Notch within Hydraulics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what v-notch describes physically and when it applies.
State every one of the 8 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: Manning's n is unitless but the constant is 1.486 (US) or 1.0 (SI).

Lecture

Why this section exists. V-Notch is the part of Hydraulics that lets you connect a channel, culvert or control structure to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as open-channel normal depth, specific energy, or a weir discharge. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. Manning's n is unitless but the constant is 1.486 (US) or 1.0 (SI). Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Row of centrifugal pumps and valved steel piping inside a water pumping station. — Photo 1. Where this shows up in practice: v-notch.
Capstone Studio instructional photograph

Hydraulics — V-Notch: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a channel, culvert or control structure. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 8 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

Q	Quantity produced by "Q = CH 5/2" — read its definition and unit from the handbook line directly above the equation.
C	Quantity produced by "C = 3.33 for rectangular weir (USCS units)" — read its definition and unit from the handbook line directly above the equation.
L	Quantity produced by "L = weir length (ft or m)" — read its definition and unit from the handbook line directly above the equation.
H	Quantity produced by "H = head (depth of discharge over weir) ft or m" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

where

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Discharge over a V-notch (triangular) weir — V-Notch

A 90° V-notch weir operates under a head of 480.0 mm with C_d = 0.595. Compute the discharge, then find the discharge if the head rises by 25% and comment on the sensitivity.

Given

Notch angle θ = 90°
H = 0.480 m
C_d = 0.595

Find

Q at H and at 1.25H

Start with the thinking

A V-notch is the meter of choice for small flows because Q varies as H^{5/2} — very sensitive at low head.
The tangent term uses half the notch angle.

Step-by-step solution

Formula
$Q = \frac{8}{15} C_{d} 2 g tan \frac{θ}{2} H^{5/2}$
Tangent
$tan (90/2) = 1.0000$
Constant
$(8/15) (0.595) (2 \times 9.81) = 1.4056$
Substituting
$Q = 1.4056 (1.0000) (0.480)^{2.5} = 0.22437 m^{3} / s$
At 1.25H
$Q = 0.39196 m^{3} / s, i . e .75$

Answer: Q = 224.4 L/s; at 1.25H it rises to 392.0 L/s

Why the other options are there

467.4 L/s (H^{3/2} used, as for a rectangular weir)
3,664,269,709,841,930,000 L/s (full angle in the tangent)

Reference: FE Reference Handbook — Hydraulics → V-Notch

Example 2

Discharge over a V-notch (triangular) weir — V-Notch (2)

A 30° V-notch weir operates under a head of 380.0 mm with C_d = 0.590. Compute the discharge, then find the discharge if the head rises by 25% and comment on the sensitivity.

Given

Notch angle θ = 30°
H = 0.380 m
C_d = 0.590

Find

Q at H and at 1.25H

Start with the thinking

A V-notch is the meter of choice for small flows because Q varies as H^{5/2} — very sensitive at low head.
The tangent term uses half the notch angle.

Step-by-step solution

Formula
$Q = \frac{8}{15} C_{d} 2 g tan \frac{θ}{2} H^{5/2}$
Tangent
$tan (30/2) = 0.2679$
Constant
$(8/15) (0.590) (2 \times 9.81) = 1.3938$
Substituting
$Q = 1.3938 (0.2679) (0.380)^{2.5} = 0.03324 m^{3} / s$
At 1.25H
$Q = 0.05807 m^{3} / s, i . e .75$

Answer: Q = 33.24 L/s; at 1.25H it rises to 58.07 L/s

Why the other options are there

87.48 L/s (H^{3/2} used, as for a rectangular weir)
71.63 L/s (full angle in the tangent)

Reference: FE Reference Handbook — Hydraulics → V-Notch

Example 3

Discharge over a V-notch (triangular) weir — V-Notch (3)

A 30° V-notch weir operates under a head of 510.0 mm with C_d = 0.595. Compute the discharge, then find the discharge if the head rises by 25% and comment on the sensitivity.

Given

Notch angle θ = 30°
H = 0.510 m
C_d = 0.595

Find

Q at H and at 1.25H

Start with the thinking

A V-notch is the meter of choice for small flows because Q varies as H^{5/2} — very sensitive at low head.
The tangent term uses half the notch angle.

Step-by-step solution

Formula
$Q = \frac{8}{15} C_{d} 2 g tan \frac{θ}{2} H^{5/2}$
Tangent
$tan (30/2) = 0.2679$
Constant
$(8/15) (0.595) (2 \times 9.81) = 1.4056$
Substituting
$Q = 1.4056 (0.2679) (0.510)^{2.5} = 0.06996 m^{3} / s$
At 1.25H
$Q = 0.12221 m^{3} / s, i . e .75$

Answer: Q = 69.96 L/s; at 1.25H it rises to 122.2 L/s

Why the other options are there

137.2 L/s (H^{3/2} used, as for a rectangular weir)
150.7 L/s (full angle in the tangent)

Reference: FE Reference Handbook — Hydraulics → V-Notch

Example 4

Discharge over a V-notch (triangular) weir — V-Notch (4)

A 30° V-notch weir operates under a head of 210.0 mm with C_d = 0.580. Compute the discharge, then find the discharge if the head rises by 25% and comment on the sensitivity.

Given

Notch angle θ = 30°
H = 0.210 m
C_d = 0.580

Find

Q at H and at 1.25H

Start with the thinking

A V-notch is the meter of choice for small flows because Q varies as H^{5/2} — very sensitive at low head.
The tangent term uses half the notch angle.

Step-by-step solution

Formula
$Q = \frac{8}{15} C_{d} 2 g tan \frac{θ}{2} H^{5/2}$
Tangent
$tan (30/2) = 0.2679$
Constant
$(8/15) (0.580) (2 \times 9.81) = 1.3702$
Substituting
$Q = 1.3702 (0.2679) (0.210)^{2.5} = 0.00742 m^{3} / s$
At 1.25H
$Q = 0.01296 m^{3} / s, i . e .75$

Answer: Q = 7.42 L/s; at 1.25H it rises to 12.96 L/s

Why the other options are there

35.33 L/s (H^{3/2} used, as for a rectangular weir)
15.99 L/s (full angle in the tangent)

Reference: FE Reference Handbook — Hydraulics → V-Notch

Example 5

Discharge over a V-notch (triangular) weir — V-Notch (5)

A 90° V-notch weir operates under a head of 460.0 mm with C_d = 0.595. Compute the discharge, then find the discharge if the head rises by 25% and comment on the sensitivity.

Given

Notch angle θ = 90°
H = 0.460 m
C_d = 0.595

Find

Q at H and at 1.25H

Start with the thinking

A V-notch is the meter of choice for small flows because Q varies as H^{5/2} — very sensitive at low head.
The tangent term uses half the notch angle.

Step-by-step solution

Formula
$Q = \frac{8}{15} C_{d} 2 g tan \frac{θ}{2} H^{5/2}$
Tangent
$tan (90/2) = 1.0000$
Constant
$(8/15) (0.595) (2 \times 9.81) = 1.4056$
Substituting
$Q = 1.4056 (1.0000) (0.460)^{2.5} = 0.20173 m^{3} / s$
At 1.25H
$Q = 0.35240 m^{3} / s, i . e .75$

Answer: Q = 201.7 L/s; at 1.25H it rises to 352.4 L/s

Why the other options are there

438.5 L/s (H^{3/2} used, as for a rectangular weir)
3,294,419,639,719,961,600 L/s (full angle in the tangent)

Reference: FE Reference Handbook — Hydraulics → V-Notch

Example 6

Discharge over a V-notch (triangular) weir — V-Notch (6)

A 90° V-notch weir operates under a head of 340.0 mm with C_d = 0.590. Compute the discharge, then find the discharge if the head rises by 25% and comment on the sensitivity.

Given

Notch angle θ = 90°
H = 0.340 m
C_d = 0.590

Find

Q at H and at 1.25H

Start with the thinking

A V-notch is the meter of choice for small flows because Q varies as H^{5/2} — very sensitive at low head.
The tangent term uses half the notch angle.

Step-by-step solution

Formula
$Q = \frac{8}{15} C_{d} 2 g tan \frac{θ}{2} H^{5/2}$
Tangent
$tan (90/2) = 1.0000$
Constant
$(8/15) (0.590) (2 \times 9.81) = 1.3938$
Substituting
$Q = 1.3938 (1.0000) (0.340)^{2.5} = 0.09395 m^{3} / s$
At 1.25H
$Q = 0.16412 m^{3} / s, i . e .75$

Answer: Q = 93.95 L/s; at 1.25H it rises to 164.1 L/s

Why the other options are there

276.3 L/s (H^{3/2} used, as for a rectangular weir)
1,534,322,584,330,741,000 L/s (full angle in the tangent)

Reference: FE Reference Handbook — Hydraulics → V-Notch

Example 7

Discharge over a V-notch (triangular) weir — V-Notch (7)

A 30° V-notch weir operates under a head of 400.0 mm with C_d = 0.595. Compute the discharge, then find the discharge if the head rises by 25% and comment on the sensitivity.

Given

Notch angle θ = 30°
H = 0.400 m
C_d = 0.595

Find

Q at H and at 1.25H

Start with the thinking

A V-notch is the meter of choice for small flows because Q varies as H^{5/2} — very sensitive at low head.
The tangent term uses half the notch angle.

Step-by-step solution

Formula
$Q = \frac{8}{15} C_{d} 2 g tan \frac{θ}{2} H^{5/2}$
Tangent
$tan (30/2) = 0.2679$
Constant
$(8/15) (0.595) (2 \times 9.81) = 1.4056$
Substituting
$Q = 1.4056 (0.2679) (0.400)^{2.5} = 0.03811 m^{3} / s$
At 1.25H
$Q = 0.06658 m^{3} / s, i . e .75$

Answer: Q = 38.11 L/s; at 1.25H it rises to 66.58 L/s

Why the other options are there

95.28 L/s (H^{3/2} used, as for a rectangular weir)
82.12 L/s (full angle in the tangent)

Reference: FE Reference Handbook — Hydraulics → V-Notch

Example 8

Discharge over a V-notch (triangular) weir — V-Notch (8)

A 90° V-notch weir operates under a head of 360.0 mm with C_d = 0.590. Compute the discharge, then find the discharge if the head rises by 25% and comment on the sensitivity.

Given

Notch angle θ = 90°
H = 0.360 m
C_d = 0.590

Find

Q at H and at 1.25H

Start with the thinking

A V-notch is the meter of choice for small flows because Q varies as H^{5/2} — very sensitive at low head.
The tangent term uses half the notch angle.

Step-by-step solution

Formula
$Q = \frac{8}{15} C_{d} 2 g tan \frac{θ}{2} H^{5/2}$
Tangent
$tan (90/2) = 1.0000$
Constant
$(8/15) (0.590) (2 \times 9.81) = 1.3938$
Substituting
$Q = 1.3938 (1.0000) (0.360)^{2.5} = 0.10838 m^{3} / s$
At 1.25H
$Q = 0.18934 m^{3} / s, i . e .75$

Answer: Q = 108.4 L/s; at 1.25H it rises to 189.3 L/s

Why the other options are there

301.1 L/s (H^{3/2} used, as for a rectangular weir)
1,770,009,661,582,712,300 L/s (full angle in the tangent)

Reference: FE Reference Handbook — Hydraulics → V-Notch

Example 9

Discharge over a V-notch (triangular) weir — V-Notch (9)

A 30° V-notch weir operates under a head of 490.0 mm with C_d = 0.585. Compute the discharge, then find the discharge if the head rises by 25% and comment on the sensitivity.

Given

Notch angle θ = 30°
H = 0.490 m
C_d = 0.585

Find

Q at H and at 1.25H

Start with the thinking

A V-notch is the meter of choice for small flows because Q varies as H^{5/2} — very sensitive at low head.
The tangent term uses half the notch angle.

Step-by-step solution

Formula
$Q = \frac{8}{15} C_{d} 2 g tan \frac{θ}{2} H^{5/2}$
Tangent
$tan (30/2) = 0.2679$
Constant
$(8/15) (0.585) (2 \times 9.81) = 1.3820$
Substituting
$Q = 1.3820 (0.2679) (0.490)^{2.5} = 0.06224 m^{3} / s$
At 1.25H
$Q = 0.10872 m^{3} / s, i . e .75$

Answer: Q = 62.24 L/s; at 1.25H it rises to 108.7 L/s

Why the other options are there

127.0 L/s (H^{3/2} used, as for a rectangular weir)
134.1 L/s (full angle in the tangent)

Reference: FE Reference Handbook — Hydraulics → V-Notch

Example 10

Discharge over a V-notch (triangular) weir — V-Notch (10)

A 90° V-notch weir operates under a head of 380.0 mm with C_d = 0.585. Compute the discharge, then find the discharge if the head rises by 25% and comment on the sensitivity.

Given

Notch angle θ = 90°
H = 0.380 m
C_d = 0.585

Find

Q at H and at 1.25H

Start with the thinking

A V-notch is the meter of choice for small flows because Q varies as H^{5/2} — very sensitive at low head.
The tangent term uses half the notch angle.

Step-by-step solution

Formula
$Q = \frac{8}{15} C_{d} 2 g tan \frac{θ}{2} H^{5/2}$
Tangent
$tan (90/2) = 1.0000$
Constant
$(8/15) (0.585) (2 \times 9.81) = 1.3820$
Substituting
$Q = 1.3820 (1.0000) (0.380)^{2.5} = 0.12302 m^{3} / s$
At 1.25H
$Q = 0.21490 m^{3} / s, i . e .75$

Answer: Q = 123.0 L/s; at 1.25H it rises to 214.9 L/s

Why the other options are there

323.7 L/s (H^{3/2} used, as for a rectangular weir)
2,009,010,613,107,197,200 L/s (full angle in the tangent)

Reference: FE Reference Handbook — Hydraulics → V-Notch

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a channel, culvert or control structure, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

V-Notch contains 8 relations; you must be able to find this page in under 15 seconds.
Exam style: open-channel normal depth, specific energy, or a weir discharge.
Unit rule: Manning's n is unitless but the constant is 1.486 (US) or 1.0 (SI).
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

Manning's n is unitless but the constant is 1.486 (US) or 1.0 (SI)
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

V-Notch

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

Q=CH5/2

Q=discharge(ft3​/secorm3​/s)

C=3.33forrectangularweir(USCSunits)

C=1.84forrectangularweir(SIunits)

C=2.54for90∘V−notchweir(USCSunits)

C=1.40for90∘V−notchweir(SIunits)

L=weirlength(ftorm)

H=head(depthofdischargeoverweir)ftorm

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section